Question

The space shuttle orbits at a distance of 335 km above the surface of the Earth. What is the gravitational force (in N) on a 1.0 kg sphere inside the space shuttle. Assume that the mass of the Earth is 5.98x1024 kg and the radius of the Earth is 6370 km


Compare (as %) the acceleration due to gravity at the altitude of the space shuttle to the local acceleration due to gravity at the surface of the Earth, 9.80 m/s2

Answer 1

the F of gravity is equal to (G*m1*m2)/(d^2) where g is the gravitational constant (6.673e-11 N*m^2/kg^2, notice the units are m not km), m1 is the mass of one object, m2 is the mass of the other object, and d= the distance between the centers of each object.

in this situation,
m1= 1.0 kg
m2 = 5.98e24 kg (i assume you meant 5.98x10^24)
d= 335 km + 6370 km. or 335000 m + 6370000 m

since it would look strange if i showed you how to put it into a calculator, i'll assume you can figure that out.

your answer is 8.87615 N, as the m^2 and 1/kg^2 in G cancel out all other units save N.

since F=mass*acceleration, and the Fgav for the object is 8.87615 and the mass is 1.0 kg, then the acceleration is equal to F/m = 8.87615/1 = 8.87615.

to get the percent, 8.87615/9.8 = x/100

the % is about 90.573%, and you would know better than I what decimal place you are expected to round it to.

Answer 2

also, if u need further help i recommend:physicsclassroom,com/Class/circles/u6l3c